4Review: combinatorial proofA combinatorial proof is a proof that uses counting arguments to prove a theorem, rather than some other method such as algebraic techniquesEssentially, show that both sides of the proof manage to count the same objectsUsually in the form of an English explanation with supporting formulae

5Polynomial expansion Consider (x+y)3: Rephrase it as:When choosing x twice and y once, there are C(3,2) = C(3,1) = 3 ways to choose where the x comes fromWhen choosing x once and y twice, there are C(3,2) = C(3,1) = 3 ways to choose where the y comes from

6Polynomial expansion Consider To obtain the x5 termEach time you multiple by (x+y), you select the xThus, of the 5 choices, you choose x 5 timesC(5,5) = 1Alternatively, you choose y 0 timesC(5,0) = 1To obtain the x4y termFour of the times you multiply by (x+y), you select the xThe other time you select the yThus, of the 5 choices, you choose x 4 timesC(5,4) = 5Alternatively, you choose y 1 timeC(5,1) = 5To obtain the x3y2 termC(5,3) = C(5,2) = 10Etc…

12Pascal’s Identity By Pascal’s identity: or 21=15+6Let n and k be positive integers with n ≥ k.Thenor C(n+1,k) = C(n,k-1) + C(n,k)The book calls this Theorem 2We will prove this via two ways:Combinatorial proofUsing the formula for

13Combinatorial proof of Pascal’s identityProve C(n+1,k) = C(n,k-1) + C(n,k)Consider a set T of n+1 elementsWe want to choose a subset of k elementsWe will count the number of subsets of k elements via 2 methodsMethod 1: There are C(n+1,k) ways to choose such a subsetMethod 2: Let a be an element of set TTwo casesa is in such a subsetThere are C(n,k-1) ways to choose such a subseta is not in such a subsetThere are C(n,k) ways to choose such a subsetThus, there are C(n,k-1) + C(n,k) ways to choose a subset of k elementsTherefore, C(n+1,k) = C(n,k-1) + C(n,k)

15(or do a Google search for ‘gulp’)April Fools Day JokesPrivacy policyFrom time to time, in order to improve Google Gulp's usefulness for our users, Google Gulp will send packets of data related to your usage of this product from a wireless transmitter embedded in the base of your Google Gulp bottle to the GulpPlex™, a heavily guarded, massively parallel server farm whose location is known only to Eric Schmidt, who carries its GPS coordinates on a 64-bit-encrypted smart card locked in a stainless-steel briefcase handcuffed to his right wrist. No personally identifiable information of any kind related to your consumption of Google Gulp or any other current or future Google Foods product will ever be given, sold, bartered, auctioned off, tossed into a late-night poker pot, or otherwise transferred in any way to any untrustworthy third party, ever, we swear. See our Privacy Policy.(or do a Google search for ‘gulp’)

18Proof practice: corollary 1Let n be a non-negative integer. ThenCombinatorial proofA set with n elements has 2n subsetsBy definition of power setEach subset has either 0 or 1 or 2 or … or n elementsThere are subsets with 0 elements, subsets with 1 element, … and subsets with n elementsThus, the total number of subsets isThus,

23Vandermonde’s identityLet m, n, and r be non-negative integers with r not exceeding either m or n. ThenThe book calls this Theorem 3

24Combinatorial proof of Vandermonde’s identityConsider two sets, one with m items and one with n itemsThen there are ways to choose r items from the union of those two setsNext, we’ll find that value via a different meansPick k elements from the set with n elementsPick the remaining r-k elements from the set with m elementsVia the product rule, there are ways to do that for EACH value of kLastly, consider this for all values of k:Thus,

25Review of Rosen, section 4.3, question 11 (a)How many bit strings of length 10 contain exactly four 1’s?Find the positions of the four 1’sThe order of those positions does not matterPositions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2Thus, the answer is C(10,4) = 210Generalization of this result:There are C(n,r) possibilities of bit strings of length n containing r ones

26Yet another combinatorial proofLet n and r be non-negative integers with r ≤ n. ThenThe book calls this Theorem 4We will do the combinatorial proof by showing that both sides show the ways to count bit strings of length n+1 with r+1 onesFrom previous slide: achieves this

27Yet another combinatorial proofNext, show the right side counts the same objectsThe final one must occur at position r+1 or r+2 or … or n+1Assume that it occurs at the kth bit, where r+1 ≤ k ≤ n+1Thus, there must be r ones in the first k-1 positionsThus, there are such strings of length k-1As k can be any value from r+1 to n+1, the total number of possibilities isThus,

28Rosen, section 4.4, question 24Show that if p is a prime and k is an integer such that 1 ≤ k ≤ p-1, then p dividesWe know thatp divides the numerator (p!) once onlyBecause p is prime, it does not have any factors less than pWe need to show that it does NOT divide the denominatorOtherwise the p factor would cancel outSince k < p (it was given that k ≤ p-1), p cannot divide k!Since k ≥ 1, we know that p-k < p, and thus p cannot divide (p-k)!Thus, p divides the numerator but not the denominatorThus, p divides

29Rosen, section 4.4, question 38Give a combinatorial proof that if n is positive integer thenProvided hint: show that both sides count the ways to select a subset of a set of n elements together with two not necessarily distinct elements from the subsetFollowing the other provided hint, we express the right side as follows:

30Rosen, section 4.4, question 38Show the left side properly counts the desired propertyConsider each of the possible subset sizes kChoosing a subset of k elements from a set of n elementsChoosing one of the k elements in the subset twice

31Rosen, section 4.4, question 38Two cases to show the right side: n(n-1)2n-2+n2n-1Pick the same element from the subsetPick that one element from the set of n elements: total of n possibilitiesPick the rest of the subsetAs there are n-1 elements left, there are a total of 2n-1 possibilities to pick a given subsetWe have to do bothThus, by the product rule, the total possibilities is the product of the twoThus, the total possibilities is n*2n-1Pick different elements from the subsetPick the first element from the set of n elements: total of n possibilitiesPick the next element from the set of n-1 elements: total of n-1 possibilitiesAs there are n-2 elements left, there are a total of 2n-2 possibilities to pick a given subsetWe have to do all threeThus, by the product rule, the total possibilities is the product of the threeThus, the total possibilities is n*(n-1)*2n-2We do one or the otherThus, via the sum rule, the total possibilities is the sum of the twoOr n*2n-1+n*(n-1)*2n-2

32Quick survey I felt I understood the material in this slide set…Very wellWith some review, I’ll be goodNot reallyNot at all

33Quick survey The pace of the lecture for this slide set was… FastAbout rightA little slowToo slow

34Quick surveyHow interesting was the material in this slide set? Be honest!Wow! That was SOOOOOO cool!Somewhat interestingRather bortingZzzzzzzzzzz